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Rouches theorem

  1. Dec 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the number of zeroes of

    p(z) = z^5 + 10z - 1

    inside |z| < 1

    2. Relevant equations



    3. The attempt at a solution

    Let f(z) = z^5
    g(z) = 10z-1

    On |z| = 1:

    10|z| - 1< |10z-1| < 10|z| + 1 (is this true...?)

    9 < |g(z)| < 11

    |f(z)| = 1

    So |g| > |f|, so f+g should have the same number of zeroes as f, which is five.

    That's incorrect obviously. What am I doing wrong?
     
  2. jcsd
  3. Dec 14, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    Set [itex]z = Re^{i\theta}[/itex]. Then
    [tex]|10z - 1|^2 = (10Re^{i\theta} - 1)(10Re^{-i\theta} - 1) = 100R^2 + 1 + 20R\cos\theta[/tex]
    so that
    [tex]
    100R^2 - 20R + 1 \leq |10z - 1|^2 \leq 100R^2 + 20R + 1
    [/tex]
    or
    [tex]
    (10R - 1)^2 \leq |10z - 1|^2 \leq (10R + 1)^2
    [/tex]
    so yes, it is true that [itex]10|z| - 1 \leq |10z-1| \leq 10|z| + 1[/itex].


    You haven't applied Rouche's Theorem correctly: if [itex]|g| > |f|[/itex] on [itex]|z| = 1[/itex] then [itex]g[/itex] and [itex]f + g[/itex] have the same number of zeroes in [itex]|z| < 1[/itex].
     
  4. Dec 14, 2013 #3
    But if |g| is between 9 and 11 and f is always 1, then f < g on |z| = 1. What's incorrect?
     
  5. Dec 14, 2013 #4

    pasmith

    User Avatar
    Homework Helper

    What's incorrect is your conclusion that f and f +g have the same number of zeroes. Rouche's Theorem is actually telling you that g and f + g have the same number of zeroes.
     
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